Coin Grab

Since we are not interested in the nature of the first coin chosen, it might as well not have been chosen - whether it is in our hand, or still in one of the boxes, does not matter. Thus we want the probability of picking a silver coin. All $16$ coins are equally likely to be picked (there is an equal chance of picking each box, and an equal choice of choosing any coin in one box, so every coin has a chance of $\tfrac14 \times \tfrac14$ of being chosen. Thus the probability of choosing a silver coin is $\tfrac{6}{16} = \tfrac38$ , making the anwer $\boxed{11}$ .

I imagined the two possible cases for drawing a second silver coin:

For the first Draw:

$P( \text{S} ) = \frac{1}{4} \left( 0 + \frac{1}{4} + \frac{1}{2} + \frac{3}{4} \right) = \frac{3}{8}$

$P( \text{G} ) = 1 - P( \text{S} ) = \frac{5}{8}$

I treated each independent case as if they happened. Thus, the probability of the second draw being silver depends on the state of the first draw.

$P ( \text{ 2nd Draw S}) = P( \text{S} ) \cdot P( \text{S} | \text{S} ) + P( \text{G} ) \cdot P( \text{S} | \text{G} )$

$P( \text{S} | \text{S} ) = \frac{0}{6} \cdot 0 + \frac{1}{6} \cdot 0 + \frac{2}{6}\cdot \frac{1}{3} + \frac{3}{6}\cdot \frac{2}{3} = \frac{4}{9}$

$P( \text{S} | \text{G} ) = \frac{4}{10} \cdot 0 + \frac{3}{10} \cdot 0 + \frac{2}{10}\cdot \frac{2}{3} + \frac{1}{10}\cdot 1 = \frac{1}{3}$

Finally:

$\begin{aligned} P ( \text{ 2nd Draw S} ) &= P( \text{S} ) \cdot P( \text{S} | \text{S} ) + P( \text{G} ) \cdot P( \text{S} | \text{G} ) \\ \quad \\ &= \frac{3}{8} \cdot \frac{4}{9} + \frac{5}{8} \cdot \frac{1}{3} \\ \quad \\ &= \frac{3}{8} \end{aligned}$

Given @Mark Hennings solution, I award myself a "C" for effort!